Question

suppose that the lengths of pregnancies are normally distributed with a mean of 269 days and a standard deviation of 16 days. use this table or the aleks calculator to find the percentage of pregnancies that are longer than 293 days. for your intermediate computations, use four or more decimal places. give your final answer to two decimal places (for example 98.23%).

Explanation:

Step1: Calculate the z - score

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $x = 293$, $\mu=269$, and $\sigma = 16$.
$z=\frac{293 - 269}{16}=\frac{24}{16}=1.5000$

Step2: Find the cumulative probability

We want $P(X>293)$. Using the standard normal distribution table, we first find $P(X\leq293)$ which corresponds to the cumulative probability for $z = 1.5000$. From the standard - normal table, $P(Z\leq1.5000)=0.9332$.

Step3: Calculate the probability of $X>293$

$P(X > 293)=1 - P(X\leq293)$. So $P(X>293)=1 - 0.9332 = 0.0668$.

Step4: Convert to percentage

To convert the probability to a percentage, we multiply by 100. So the percentage is $0.0668\times100 = 6.68\%$.

Answer:

$6.68\%$